void main(){
double fVal = 66.6f;
int rate = 10;

int intValue = cast(int)(fVal*rate);
printf("Value = %d\n", intValue);
}

// Value = 665
// What is the matter?

On Wed, 17 Nov 2004 09:35:52 +0000 (UTC), LiuXuHong <LiuXuHong_member@pathlink.com> wrote:

> void main(){
> double fVal = 66.6f;
> int rate = 10;
>
> int intValue = cast(int)(fVal*rate);
> printf("Value = %d\n", intValue);	
> }
>
> // Value = 665
> // What is the matter?
>
>

This is what is most likely happing:
fVal : 66.6
x10  : ~666.0 (ie 665.93853...)
(int): 665 [ .93853 cut off]

This is due to the fact that floating points are represented in binary,
and therefore only values like xxx/1024 can be represented exactly,
and when a floating point number is cast to an int, everything after the
decemal place is simply truncated.
If you must round the number properly, add .5 to it before casting.

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On Thu, 18 Nov 2004 01:09:13 +1300, Simon Buchan <currently@no.where> wrote:
> On Wed, 17 Nov 2004 09:35:52 +0000 (UTC), LiuXuHong  <LiuXuHong_member@pathlink.com> wrote:
>
>> void main(){
>> double fVal = 66.6f;
>> int rate = 10;
>>
>> int intValue = cast(int)(fVal*rate);
>> printf("Value = %d\n", intValue);	
>> }
>>
>> // Value = 665
>> // What is the matter?
>>
>>
>
> This is what is most likely happing:
> fVal : 66.6
> x10  : ~666.0 (ie 665.93853...)
> (int): 665 [ .93853 cut off]
>
> This is due to the fact that floating points are represented in binary,
> and therefore only values like xxx/1024 can be represented exactly,
> and when a floating point number is cast to an int, everything after the
> decemal place is simply truncated.
> If you must round the number properly, add .5 to it before casting.

Depending on the result you're after you can also move the cast i.e.

to get 666: intValue = (cast(int)fVal*rate);
(this truncates the 66.6 to 66, then does x10)

to get 670: intValue = (cast(int)(fVal+0.5)*rate);
(this does what Simon mentioned, then truncates, then does x10)

Regan

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On Thu, 18 Nov 2004 11:39:19 +1300, Regan Heath <regan@netwin.co.nz> wrote:

 <snip>
> Depending on the result you're after you can also move the cast i.e.
>
> to get 666: intValue = (cast(int)fVal*rate);
> (this truncates the 66.6 to 66, then does x10)
>
> to get 670: intValue = (cast(int)(fVal+0.5)*rate);
> (this does what Simon mentioned, then truncates, then does x10)
>
> Regan
>

Actually the top one will give you 660 (since when does 66 x 10 give 666 :D),
and the bottom will give you 671. (67.1 x 10)

You only multiply by .5 just before the cast, of course. (cast(int)(fVal*rate + 0.5))
This gives you your 666.

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On Thu, 18 Nov 2004 15:13:59 +1300, Simon Buchan <currently@no.where> wrote:
> On Thu, 18 Nov 2004 11:39:19 +1300, Regan Heath <regan@netwin.co.nz> wrote:
>
>   <snip>
>> Depending on the result you're after you can also move the cast i.e.
>>
>> to get 666: intValue = (cast(int)fVal*rate);
>> (this truncates the 66.6 to 66, then does x10)
>>
>> to get 670: intValue = (cast(int)(fVal+0.5)*rate);
>> (this does what Simon mentioned, then truncates, then does x10)
>>
>> Regan
>>
>
> Actually the top one will give you 660 (since when does 66 x 10 give 666  :D),

Doh! I mean't to type that ;o)

> and the bottom will give you 671. (67.1 x 10)

No, it gives 670. The cast to int truncates 67.1 to 67 before the x10 occurs.
(cast must have higher precedence than *?)

> You only multiply by .5 just before the cast, of course.  (cast(int)(fVal*rate + 0.5))
> This gives you your 666.

Ahh, so that is what you meant. Multiply, add 0.5 then cast to int truncating 666.x to 666.

Regan

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